Problem: $\begin{aligned} &F(x)=\sqrt{x+7} \\\\ &f(x)=F'(x) \end{aligned}$ $\int_{2}^{9} f(x)\,dx=$
Explanation: $f$ is the derivative of $F$, which means $F$ is an antiderivative of $f$. Since we know the antiderivative of $f$, we can use the fundamental theorem of calculus: For every function $f$ and its antiderivative $F$, $\int_a^b f(x)\,dx=F(b)-F(a)$. $\begin{aligned} &\phantom{=}\int_{2}^{9} f(x)\,dx \\\\ &=F({9})-F({2}) \\\\ &=\sqrt{{9}+7}-\sqrt{{2}+7} \\\\ &=4-3 \\\\ &=1 \end{aligned}$ In conclusion, $\int_{2}^{9} f(x)\,dx=1$